Limit of the difference between upper and lower sums - Proving integrability Let $f$ be a bounded function on some interval $[a,b]$.
Suppose there exists a sequence of partitions $(D_n)^\infty _{n=1}$ of that interval and that:
$$lim_{n->\infty}{}[U(f,D_n)-L(f,D_n)] = 0$$
How would we prove $f$ is integrable?
I thought about the following:
As $f$ is bounded on the interval then its upper and lower sums are also bounded so they converge. (How would you say it formally? is it trivial to say it?)
Then we would get that $S_u=S_l$ (notation for the corresponding finite limit)
And then we get that the supremum of the lower sum is equal to the infimum of the upper sum which imply that $f$ is integrable.
What do you think?
 A: The integrability Riemann's criterion say:
A bounded function $f : [a,b]\to\mathbb R$ is Riemann integrable if and only if:
$$\forall \epsilon>0\;\; \exists \text{ a partition }\; D \;\text{ of }\; [a,b]\; \text{ such that }\; U(f; D)-L(f;D)\leq\epsilon. $$
Now, by the limit in hypothesis, for any fixed $\epsilon>0$, there exists $\nu\in\mathbb N$ such that
$$ n\geq\nu \implies U(f;D_n)-L(f;D_n)\leq\epsilon, $$
and therefore $f$ is integrable.
Conversely, let $f\,$ be integrable. Taking, for any $n\in\mathbb N,\;$ $\epsilon = \frac 1 n$, we deduce, still by the Riemann criterion, that there exists a partition $D_n$ of $[a,b]$ such that
$$ U(f;D_n)-L(f;D_n)\leq1/n, $$
so we have
$$\lim_{n\to\infty\;}[U(f;D_n)-L(f;D_n)]=0. $$
A: Assuming $D_{i+1}$ is a refinement of $D_i$.
The Riemann sum $\sigma$ of $f$ over a partition $D$ satisfies $$L(D)\le\sigma\le U(D)$$
Since $f$ is bounded on $[a,b]$, $U$ and $L$ are bounded on $[a,b]$.
Since $U$ and $L$ are monotone, they converges.
Hence $\sigma$ converges.
(However it is not very clear what properties $D$ has. So it is not yet possible to show that $f$ is integrable. Counter example: $[a,b]=[0,1]$; $D_1=\{[0,\frac 1 2],[\frac 1 2,1]\}$, $D_2=\{[0,\frac 1 4],[\frac 1 4,\frac 1 2],[\frac 1 2,1]\}$, $D_3=\{[0,\frac 1 8],[\frac 1 8,\frac 1 4],[\frac 1 4,\frac 1 2],[\frac 1 2,1]\}$...)
